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            8051TimerProgramming inAssembly andC                      Microcontroller
                         8051 Timer Programming in Assembly and C
            Objectives:
            At the end of this chapter, we will be able to:
                 List the timers of 8051 and their associated registers
                 Describe the various modes of the 8051 timers.
                 Program the 8051 timers in assembly and C and
                 Program the 8051 counters in assembly and C.
            Programming 8051 Timers:
            The 8051 has two timers/counters; they can be used either as Timers to generate a time delay
            or as event counters to count events happening outside the microcontroller.
            Basic Timers of 8051:
            Both Timer 0 and Timer 1 are 16 bits wide. Since 8051 has an 8-bit architecture, each 16-bits
            timer is accessed as two separate registers of low byte and high byte. The low byte register is
            called  TL0/TL1 and the  high  byte  register  is  called  TH0/TH1.  These  registers  can  be
            accessed like any other register. For example:
                 MOV TL0,#4FH
                 MOV R5,TH0
                                    Figure 1: Timer 0 and Timer1 register
            TMOD (timer mode) Register:
            Both timers 0 and 1 use the same register, called TMOD (timer mode), to set the various
            timer operation modes. TMOD is an 8-bit register. The lower 4 bits are for Timer 0 and the
            upper 4 bits are for Timer 1. In each case, the lower 2 bits are used to set the timer mode the
            upper 2 bits to specify the operation. The TMOD register is as shown in figure 2 below:
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            8051TimerProgramming inAssembly andC                      Microcontroller
            Timers of 8051 do starting and stopping by either software or hardware control. In using
            software to start  and stop the timer where GATE=0. The start and stop of the timer are
            controlled by way  of  software  by  the  TR  (timer  start)  bits  TR0 and  TR1. The  SETB
            instruction starts it, and it is stopped by the CLR instruction. These instructions start and stop
            the timers as long as GATE=0 in the TMOD register. The hardware way of starting and
            stopping the timer by an external source is achieved by making GATE=1 in the TMOD
            register.
            Mode 1 Programming:
            The following are the characteristics and operations of mode 1:
            1.  It  is  a  16-bit  timer; therefore, it  allows value of 0000 to FFFFH to be loaded into the
            timer€s register TL and TH.
            2. After TH and TL are loaded with a 16-bit initial value, the timer must be started. This is
            done by SETB TR0 for timer 0 and SETB TR1 for timer 1.
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            8051TimerProgramming inAssembly andC                      Microcontroller
            3. After the timer is started, it starts to count up. It counts up until it reaches its limit of
            FFFFH. When it rolls over from FFFFH to 0000, it sets high a flag bit called TF (timer flag).
            Each timer has its own timer flag: TF0 for timer 0 and TF1 for timer 1. This timer flag can be
            monitored. When this timer flag is raised, one option would be to stop the timer with the
            instructions CLR TR0 or CLR TR1, for timer 0 and timer 1, respectively.
            4. After the timer reaches its limit and rolls over, in order to repeat the process. TH and TL
            must be reloaded with the original value, and TF must be reloaded to 0.
            Steps to program in mode 1:
            To generate a time delay, using timer in mode 1, following are the steps:
            1. Load the TMOD value register indicating which timer (timer 0 or timer 1) is to be used and
            which timer mode (0 or 1) is selected.
            2. Load registers TL and TH with initial count value.
            3. Start the timer.
            4. Keep monitoring the timer flag (TF) with the JNB TFx, target instruction to see if it is
            raised. Get out of the loop when TF becomes high.
            5. Stop the timer.
            6. Clear the TF flag for the next round.
            7. Go back to Step 2 to load TH and TL again.
             Example 1
             In the following program, we create a square wave of 50% duty cycle (with equal portions
             high and low) on the P1.5 bit. Timer 0 is used to generate the time delay. Analyze the
             program. Also calculate the delay generated. Assume XTAL=11.0592MHz.
             Program:
                  MOV TMOD,#01      ;Timer 0, mode 1(16-bit mode)
             HERE: MOV TL0,#0F2H    ;TL0=F2H, the low byte
                  MOV TH0,#0FFH     ;TH0=FFH, the high byte
                  CPL P1.5          ;toggle P1.5
                  ACALL DELAY
                  SJMP HERE
             DELAY:
                  SETB TR0          ;start the timer 0
             AGAIN: JNB TF0,AGAIN ;monitor timer flag 0 until it rolls over
                  CLR TR0           ;stop timer 0
                  CLR TF0           ;clear timer 0 flag
                  RET
             (a)In the above program notice the following step.
             1. TMOD is loaded.
             2. FFF2H is loaded into TH0-TL0.
            Prof. Roopa Kulkarni, GIT, Belgaum                                   Page3
             3. P1.5 is toggled for the high and low portions of the pulse.
             4. The DELAY subroutine using the timer is called.
             5. In the DELAY subroutine, timer 0 is started by the SETB TR0 instruction.
             6. Timer 0 counts up with the passing of each clock, which is provided by the crystal
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            8051TimerProgramming inAssembly andC                      Microcontroller
             Example 2: Find the delay generated by timer 0 in the following code, using hex as well
             as decimal method. Do not include the overhead due to instruction.
             Program:
                   CLR P2.3          ;Clear P2.3
                   MOV TMOD,#01      ;Timer 0, 16-bitmode
             HERE: MOV TL0,#3EH      ;TL0=3Eh, the low byte
                   MOV TH0,#0B8H     ;TH0=B8H, the high byte
                   SETB P2.3         ;SET high timer 0
                   SETB TR0          ;Start the timer 0
             AGAIN: JNB TF0,AGAIN ;Monitor timer flag 0
                   CLR TR0           ;Stop the timer 0
                   CLR TF0           ;Clear TF0 for next round
                   CLR P2.3
             Solution: (a) (FFFFH – B83E + 1) = 47C2H = 18370 in decimal and 18370 × 1.085 us =
             19.93145 ms
             (b) Since TH – TL = B83EH = 47166 (in decimal) we have 65536 – 47166 = 18370. This
             means that the timer counts from B38EH to FFFF. This plus Rolling over to 0 goes
             through  a  total  of  18370  clock cycles,  where  each  clock  is  1.085µs  in  duration.
             Therefore, we have 18370 × 1.085 us = 19.93145 ms as the width of the pulse.
            Finding values to be loaded into the timer:
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